\chapter{派生观测值平差和分步平差(仅限观测值)}

\section{引言}

In earlier discussions it was shown that observational variables need not be the direct numerical outcomes of the physical operations of measurement.\\
在之前的讨论中表明观测变量不需要是实际测量操作的直接数值结果。

In fact, as was shown by Example 7.5, we could use in the adjustment other stochastic vaiables that are derived from the originally available variables.\\
事实上，正如示例 7.5 所示，我们可以在平差中使用其他从最初可用的变量派生的随机变量。

The outcome is exactly the same provided the stochastic model is modified in such a way that the stochastic properties of the derived observations are also properly obtained. \\
%这一句待定，没太懂意思

This rather important concept will be discussed further and demonstrated by examples in the first part of this chapter.\\
这个非常重要的概念将进一步讨论，并在本章第一部分的例子中得到说明。

The second part of the chapter will deal with another related concept.\\
本章的第二部分将讨论另一个相关的概念。

It involves what is termed “adjustment in steps” when the technique applied is that of adjusting observations only. \\
它涉及被称为"分步平差"，当技术只被用于观测值平差。

In many practical situations a set of observations is obtained that belongs to a certain (geometrical) model. \\
在很多实际情况下，获得一组属于特定（几何）模型的观测结果。

This first set is then adjusted and estimates for the observations are obtained from the adjustment. \\
然后对第一组平差，从平差中获得观测值的估计值。

At a later stage, a second set of observations may be acquired that belongs to either the same model or to an expanded form of it. \\
在稍后阶段，可以获取属于同一模型或该模型的扩展形式的第二组观测值。

Thus the adjustment of the second step would involve in it not only the second set of observations but also some or all of the new estimates of the observations in the first set.\\
因此，第二步的平差将不仅涉及第二组观测结果，而且涉及第一组观测结果的一些或全部新的估计数。

By necessity, then, the adjustment is performed in two steps and the final result must be the same as would have been obtained had we performed the total adjustment of all sets of observations together. This will be demonstrated both algebraically and by numerical example.\\
因此，平差必须分两个步骤进行，最终结果必须与我们同时执行所有观测结果的总平差结果相同。

\section{用导出的观测值进行平差}

Consideration is limited to linear functions (which may be originally linear or may have been linearized). \\
考虑仅限于线性函数（可能最初是线性函数，也可能已线性化）。

Also only the case of adjustment of observations is considered here, whereas other possibilities, which include parameters, will be treated in Part III of this book.\\
这里也仅仅考虑观测值平差，而其他可能性包括参数将在本书的第三部分中处理。

With $n$ observations and $r$ redundancy the conditions may be written as\\
使用$n$个观测值和$r$个多余观测量，可以写出
\begin{equation}
    \underset{r, n }{\mathbf{A}_{l}} \underset{n, 1} {\mathbf{v}_{l}}=\left(-\mathbf{A}_{l} \boldsymbol{l}+\mathbf{d}_{l}\right)=\underset{r, 1}{\mathbf{f}_{l}}\label{eq:10-1}
\end{equation}
With $d_l$  and $f_l$  being constant term vectors.\\
$d_l$和$f_l$是常量向量。

Associated with$l$is the cofactor matrix$Q_{ll}$. \\
与$l$关联的是余因数矩阵$Q_{ll}$。

If an adjustment is carried out using these conditions, the following relations may be written (refer to Chapter 7)\\
如果使用这些条件进行平差，可以写出以下关系(请参阅第 7 章):
$$
    \mathbf{k}_{l}=\left(\mathbf{A}_{l} \mathbf{Q}_{l l} \mathbf{A}^{t}_{l}\right)^{-1} \mathbf{f}_{l} = \mathbf{W}_{e l} \mathbf{f}_{l}
$$

$$
    \mathbf{v}_{l} = \mathbf{Q}_{l l} \mathbf{A}^{t}_{l} \mathbf{k}_{l}= \mathbf{Q}_{l l} \mathbf{A}^{t}_{l} \mathbf{W}_{e l} \mathbf{f}_{l}
$$

$$
    \mathbf{Q}_{v l v l} = \mathbf{Q}_{l l} \mathbf{A}^{t}_{l} \mathbf{W}_{e l} \mathbf{A}_{l} \mathbf{Q}_{ll}
$$

$$
    \mathbf{Q}_{l l} = \mathbf{Q}_{l l} - \mathbf{Q}{ v l v l }
$$
Suppose now that we are to derive a new set of observations $m_{p,1}$  from the original set $l$ , such that $p \leq n$. \\
假设现在我们要从原始集$\bm{l}$获得一组新的观测值$\bm{m}_{p,1}$以便$p \leq n$。

This new set can be used in the adjustment in place of the original set with results consistent with those that would have been obtained from $\bm{l}$, provided the following assertions are considered:\\
这个新集可用于平差，以代替原始集，结果与从$l$中获得的结果一致，前提是考虑以下断言：

\begin{enumerate}
    \item The number of derived observations is less than or equal to the number of original observations, that is, $p \leq n$.\\
          派生观测值的数量小于或等于原始观测值的数量，就是$p \leq n$
    \item The cofactor matrix $Q{mm}$ for the new set m must be properly evaluated.\\
          必须正确评估新集 m 的余因数矩阵$Q{mm}$。
          Thus if\\
          因此，如果
          \begin{equation}
              \underset{p,1}{\mathbf{m}} =  \underset{p,n}{\mathbf{D}} \underset{n,l}{\boldsymbol{l}} \text{and} \mathbf{v}_{m} = \mathbf{D} \mathbf{v}_{l}\label{eq:10-2}
          \end{equation}
          Then
          \begin{equation}
              \mathbf{Q}_{m m} = \mathbf{D} \mathbf{Q}_{l l} \mathbf{D}^{t}\label{eq:10-3}
          \end{equation}
          must be computed.\\
          必须计算.

    \item Redundancy, or the number of degrees of freedom r, must remain the same after obtaining the derived observations. In other words the model must not change while switching over from one set of observations to another if we are to get consistent adjustment results.\\
          多余观测量值或自由度 r 在获得派生观测值后必须保持不变。换句话说，如果我们要获得一致的平差结果，模型在从一组观测值切换到另一组观测值时不得更改。

\end{enumerate}
Because of point (1) above, we may write the conditions in terms of the derived observations m as follows:\\
由于上述第 （1） 点，我们可以根据派生观测值 m 编写条件，如下所示:
\begin{equation}
    \mathbf{A}_{m} \mathbf{v}_{m} = \left(-\mathbf{A}_{m} \boldsymbol{m} + \mathbf{d}_{m} \right) = \mathbf{f}_{m} \label{eq:10-4}
\end{equation}
using a vector $f_m$ that is different from $f_l$ , allows for the possibility of using a different set of conditions, even though the number of conditions is the same. For the obvious reason that any r conditions chosen are independent, then\\
使用与 B 不同的向量 A，允许使用一组不同的条件，即使条件的数量相同。显然，选择的任何 r 条件都是独立的，然后
\begin{equation}
    \mathbf{f}_{m} = \mathbf{C} \mathbf{f}_{l}\label{eq:10-5}
\end{equation}
with $C$ being nonsingular.\\
$C$ 是非奇异的。

Applying least squares to equation (10.4) leads directly to the following relations:\\
对方程 \eqref{eq:10-4} 应用最小二乘直接得到以下关系：
\begin{subequations}
    \begin{align}
        \mathbf{k}_{m}               & =\mathbf{W}_{e_{m}} \mathbf{f}_{m}=\left(\mathbf{A}_{m} \mathbf{Q}_{m m} \mathbf{A}_{m}^{t}\right)^{-1} \mathbf{f}_{m}\label{eq:10-6a}    \\
        \mathbf{v}_{m}               & =\mathbf{Q}_{m m} \mathbf{A}_{m}^{t} \mathbf{k}_{m}=\mathbf{Q}_{m m} \mathbf{A}_{m}^{t} \mathbf{W}_{e_{m}} \mathbf{f}_{m}\label{eq:10-6b} \\
        \mathbf{Q}_{v_{m} v_{m}}     & =\mathbf{Q}_{m m} \mathbf{A}_{m}^{t} \mathbf{W}_{e_{m}} \mathbf{A}_{m} \mathbf{Q}_{m m}\label{eq:10-6c}                                   \\
        \mathbf{Q}_{\hat{m} \hat{m}} & =\mathbf{Q}_{m m}-\mathbf{Q}_{v_{m} v_{m}}\label{eq:10-6d}
    \end{align}\label{eq:10-6}
\end{subequations}

Equations (10.6) give the least squares solution in terms of the new set of observations m and their a priori cofactor matrix Qmm.\\
方程\eqref{eq:10-6}根据新的观测值$m$及其先验余因数矩阵$Q_{mm}$，给出最小二乘解。

We must now compute the residuals   and the a posteriori matrix Qvlvl, from these, and also prove that they are the same values that would have been obtained if   and   were used directly.\\
我们现在必须计算残差$v_l$和这些残差矩阵$Q_{vlvl}$，并证明它们与直接使用$l$和$Q_{ll}$时获得的值相同

From equations \eqref{eq:10-1}, \eqref{eq:10-2}, \eqref{eq:10-4}, and \eqref{eq:10-5} it follows that：\\
从方程\eqref{eq:10-1}，\eqref{eq:10-2}，\eqref{eq:10-4}和 \eqref{eq:10-5}，得如下：
$$
    \mathbf{A}_{m} \mathbf{D} \mathbf{v}_{l}=\mathbf{C} \mathbf{f}_{l}
$$

$$
    \left(\mathbf{C}^{-1} \mathbf{A}_{m} \mathbf{D} \right) \mathbf{v}_l = \mathbf{f}_l
$$
which leads to\\
可得出
\begin{equation}
    \mathbf{A}_{l} = \mathbf{C}^{-1} \mathbf{A}_{m} \mathbf{D}\label{eq:10-7}
\end{equation}
From equations (7.4, 10.3, 10.5, and 10.6a) we write\\
从方程（\eqref{eq:7-4}，\eqref{eq:10-3}，\eqref{eq:10-5}和\eqref{eq:10-6a}）我们可以写出
$$
    \begin{aligned}
        \mathbf{k}_{l} & =\left(\mathbf{A}_{l} \mathbf{Q}_{l l} \mathbf{A}^{t}_{l} \right)^{-1} \mathbf{f}_{l}                       \\
                       & =[\left(\mathbf{C}^{-1} \mathbf{A}_{m} \mathbf{D} \right) \mathbf{Q}_{l l} \left( \mathbf{D}^{t} \mathbf{A}_{m} (\mathbf{(C}^{-1})^{t} \right)]^{-1} \mathbf{f}_{l} \\
                       & =[\mathbf{C}^{-1} \mathbf{A}_{m} \mathbf{Q}_{m m} \mathbf{A}^{t}_{m} (\mathbf{C}^{t})^{-1} \mathbf{f}_{l}             \\
                       & =\mathbf{C}^{t} \left(\mathbf{A}_{m} \mathbf{Q}_{m m} \mathbf{A}^{t}_{m} \right)^{-1} \mathbf{C} \mathbf{f}_{l}                   \\
                       & =\mathbf{C}^{t} \mathbf{W}_{e_m} \mathbf{f}_{m}
    \end{aligned}
$$
Or
\begin{equation}
    \mathbf{k}_{l} = \mathbf{C}^{t} \mathbf{k}_{m}
\end{equation}\label{eq:10-8}

Now, to derive a relation for $v_l$  in terms of $v_m$ ,\\
现在，要推导出$v_l$ 的关系，以$v_m$
$$
    \begin{aligned}
        \mathbf{v}_l & =\mathbf{Q}_{l l} \mathbf{A}^{t}_{l} \mathbf{k}_{l}              \\
            & =\mathbf{Q}_{l l} \mathbf{D}^{t} \mathbf{A}^{t}_{m} (\mathbf{C}^{-1})^{t} \mathbf{C}^{t} \mathbf{k}_{m}
    \end{aligned}
$$
Or
\begin{subequations}
    \begin{align}
    \mathbf{v}_{l} = \mathbf{Q}_{l l} \mathbf{D}^{t} \mathbf{A}^{t}_{m} \mathbf{k}_{m}  \label{eq:10-9a}\\
    \mathbf{v}_{l} = \mathbf{Q}_{l l} \mathbf{D}^{t} \mathbf{Q}^{-1}_{m m} \mathbf{v}_{m} \label{eq:10-9b}
    \end{align} \label{eq:10-9}
\end{subequations}
It should be straightforward to show that premultiplying both sides of equation (10.9b) by D yields equation (10.2), taking equation (10.3) into account.\\
应该直接显示，将公式\eqref{eq:10-9b}两边乘以D产生公式\eqref{eq:10-2}，同时考虑公式\eqref{eq:10-3}。

Using equation (10.9b) we can derive a relationship for the a posteriori cofactor matrix  .\\
使用方程\eqref{eq:10-9b}，我们可以推导出后余因数矩阵$Q_{vlvl}$的关系。
$$
    \mathbf{Q}_{vlvl} = \left( \mathbf{q}_{l l} \mathbf{D}^{t} \mathbf{Q}^{-1}_{mm} \right) \mathbf{Q}_{v_m v_m} \left( \mathbf{Q}_{l l} \mathbf{D}^{t} \mathbf{Q}^{-1}_{m m} \right)^{t}
$$
And from equation (10.6c),\\
从方程\eqref{eq:10-6c},
$$
    \mathbf{Q}_{v_l v_l} = \mathbf{Q}_{l l} \mathbf{D}^{t} \mathbf{Q}^{-1}_{mm} \mathbf{Q}_{m m} \mathbf{A}^{t}_{m} \mathbf{W}_{e_m} \mathbf{A}_{m} \mathbf{Q}_{m m} \mathbf{Q}^{-1}_{m m} \mathbf{D} \mathbf{Q}_{l l}
$$
Or
\begin{equation}
    \mathbf{Q}_{v_l v_l} = \mathbf{Q}_{l l} \mathbf{D}^{t} \mathbf{A}^{t}_{m} \mathbf{W}_{e_m} \mathbf{A}_{m} \mathbf{D} \mathbf{Q}_{l l} \label{eq:10-10}
\end{equation}
Finally, it must be shown that equation (10.10) is equivalent to equation (7.7).\\
最后，必须明确方程（10.10）等效于方程（7.7）。
$$
    \mathbf{Q}_{v v_{l}}=\mathbf{Q}_{l l} \mathbf{D}^{t} \mathbf{A}_{m}^{t}\left(\mathbf{A}_{m} \mathbf{D} \mathbf{Q}_{l l} \mathbf{D}^{t} \mathbf{A}_{m}^{t}\right)^{-1} \mathbf{A}_{m} \mathbf{D} \mathbf{Q}_{\boldsymbol{u}}
$$
and from equation (10.7),\\
从方程\eqref{eq:10-7},
$$
    \begin{aligned}
        \mathbf{Q}_{v l v l} & = \mathbf{Q}_{l l} \mathbf{D}^{t} \mathbf{A}^{t}_{m} \left( \mathbf{A}_{m} \mathbf{D} \mathbf{Q}_{l l} \mathbf{D}^{t} \mathbf{A}^{t}_{m} \right)^{-1} \mathbf{A}_{m} \mathbf{D} \mathbf{Q}_{l l}               \\
                             & = \mathbf{Q}_{l l} [ \mathbf{D}^{t} \mathbf{A}^{t}_{m} (\mathbf{C}^{-1})^{t}] \left( \mathbf{A}_{l} \mathbf{Q}_{l l} \mathbf{A}^{t}_{l} \right)^{-1} \left( \mathbf{C}^{-1} \mathbf{A}_{m} \mathbf{D} \right) \mathbf{Q}_{l l}
    \end{aligned}
$$
or
$$
    \mathbf{Q}_{v l v l} = \mathbf{Q}_{l l} \mathbf{A}^{t}_{l} \mathbf{W}_{e l} \mathbf{A}^{t}_{l} \mathbf{Q}_{l l}
$$
which is identical to equation (7.7). To show that\\
与方程 \eqref{eq:7-7} 相同。表示
$$
    \mathbf{v}^{t}_{l} \mathbf{W}_{l l} \mathbf{v}_{l} = \mathbf{v}^{t}_{m} \mathbf{W}_{m m} \mathbf{v}_{m}
$$
We refer to equations (10.9b) and (10.3).\\
我们参见方程\eqref{eq:10-9b}和\eqref{eq:10-3}
$$
    \begin{aligned}
        \mathbf{v}^{t}_{l} \mathbf{W}_{l l} \mathbf{v}_{l} & =\left( \mathbf{v}^{t}_{m} \mathbf{Q}^{-1}_{m m} \mathbf{D} \mathbf{Q}_{l l} \right) \mathbf{W}_{l l} \left( \mathbf{Q}_{l l} \mathbf{D}^{t} \mathbf{Q}^{-1}_{m m} \mathbf{v}_{m} \right) \\
                                                           & =\mathbf{v}^{t}_{m} \mathbf{q}^{-1}_{m m} \left( \mathbf{D} \mathbf{Q}_{l l} \mathbf{D}^{t} \right) \mathbf{Q}^{-1}_{m m} \mathbf{v}_{m}               \\
                                                           & =\mathbf{v}^{t}_{m} \mathbf{Q}^{-1}_{m m} \mathbf{v}_{m} = \mathbf{v}^{t} \mathbf{W}_{m m} \boldsymbol{v}
    \end{aligned}
$$
To demonstrate the relations derived above some munerical examples follow.\\
为了说明上述一些具体例子得出的关系。

Example 10.1.As a first example consider the simple case of a plane triangle ABC in Figure 10.1. \\
示例 10.1.作为第一个示例，请考虑图 10.1 中平面三角形 ABC 的简单情况。

Six directions are measured, two at each corner.\\
测量六个方向，每个角两个方向。

From these six directions, which may be considered as the original observations, three angles may be computed as derived observations.\\
从这六个方向可视为原始观测值的方向，可以计算三个角度作为派生观测值。

It is required to show that from the adjustment with the derived observations (angles), residuals for the original directions may be computed as well as their cofactor matrix which are the same as those obtained from adjustment with the directions themselves.\\
需要表明，从与派生观测值（角度）的平差中，可以计算原始方向的残差及其余因数矩阵，这些矩阵与从平差中获得的方向本身相同。

Assume that the cofactor matrix for the directions is the identity matrix and that the triangle’s misclosure (contradiction or discrepancy) is w seconds of arc.\\
假设方向的余因数矩阵是标识矩阵，并且三角形的闭合差是圆弧的 w 秒。

\begin{figure}[H]
    \centering
    \includegraphics{images//10-1.png}
    \caption{ }
    \label{img:10-1}
\end{figure}

Solution: For the sake of later checks, the triangle is first adjusted using directions.\\
解决方案：为了以后的检查，首先使用方向调整三角形。

Since there are three zero directions, one at each corner, and since two variables are necessary for determining the shape of the triangle, then r = 1. \\
由于有三个零方向，每个角一个，并且由于两个变量是确定三角形的形状所必需的，然后 r = 1。

This is the same as adjusting by angles, which satisfies requirement (3) in Section 10.2. \\
这与按角度进行调整相同，满足第 10.2 节中的要求 （3）。

From this it is obvious that there is only one f, or   which leads to C = I, in equation (10.5).\\
从这里开始，很明显，在方程\eqref{eq:10-5}中，只有一个$f$或$f_l=f_m=w$导致$C=I$。

The condition for adjustment by directions is\\
按方向的条件平差为
$$
    \left[\begin{array}{lllllll}
            -1 & 1 & -1 & 1 & -1 & 1
        \end{array}\right] \mathbf{v}_{d}=w \quad \text { or } \quad \underset{1,6}{\mathbf{A}_{d}} \underset{\mathbf{6}, 1}{\mathbf{v}_{d}}=\underset{1,1}{\mathbf{f}}
$$
With
$$
    \mathbf{Q}_{\mathrm{d} d}=\mathbf{I} \quad \mathbf{Q}_{e_{d}}=6 \quad k_{d}=k=\frac{w}{6} \quad \mathbf{v}_{d}^{t}=\left[\begin{array}{llllll}
            -1 & 1 & -1 & 1 & -1 & 1
        \end{array}\right] \frac{\boldsymbol{w}}{6}
$$

$$
    \mathbf{Q}_{v d v d}=\frac{1}{6}
    \left[\begin{array}{rrrrrr}
            1 & -1 & 1  & -1 & 1  & -1 \\
              & 1  & -1 & 1  & -1 & 1  \\
              &    & 1  & -1 & 1  & -1 \\
              &    &    & 1  & -1 & 1  \\
              &    &    &    & 1  & -1 \\
              &    &    &    &    & 1
          \end{array}\right]
$$
To get the angles\\
获取角度
$$
    \left[\begin{array}{r}
            \boldsymbol{l}_1\\
            \boldsymbol{l}_2\\
            \boldsymbol{l}_3
          \end{array}\right]
           = 
    \left[\begin{array}{rrrrrr}
            -1&1&0&0&0&0\\
            0&0&-1&1&0&0\\
            0&0&0&0&-1&1
          \end{array}\right]    
    \left[\begin{array}{r} 
        \boldsymbol{d}_{10}\\
        \boldsymbol{d}_{11}\\
        \boldsymbol{d}_{12}\\
        \boldsymbol{d}_{13}\\
        \boldsymbol{d}_{14}\\
        \boldsymbol{d}_{15}
          \end{array}\right]  \quad  \text{or}  \quad  \boldsymbol{l}=\mathbf{D} \boldsymbol{d}
$$
and
$$
    \mathbf{Q}_{l l} = \mathbf{D} \mathbf{Q}_{d d} \mathbf{D}^{t} = 2\mathbf{I}_{3}
$$
The condition equation is\\
条件方程是
$$
    \left[\begin{array}{lll}
            1&1&1
          \end{array}\right]  \mathbf{v}_{t} = \mathbf{w}
$$

$$
    \mathbf{Q}_{e {m}} = 6   \quad    \mathbf{k}_{l} = \mathbf{k} = \frac{w}{6}
$$

$$
    \mathbf{V}^{t}_{l} = \frac{w}{3} 
    \left[\begin{array}{rrr}
            1&1&1
          \end{array}\right]
$$

$$
    \mathbf{Q}_{v_l v_l} = \frac{2}{3} 
    \left[\begin{array}{rrr}
            1&1&1\\
             &1&1\\
             & &1
          \end{array}\right]
$$
Now compute   and from , and other matrices belonging to the adjustment. From equation (10.9a),\\
现在从属于平差的$v_l$,$Q_{vlvl}$和其他矩阵计算$v_d$和$Q_{vdvd}$。从方程\eqref{eq:10-9a}，
$$
    \mathbf{v}_{d} = \mathbf{Q}_{d d} \mathbf{D}^{t} \mathbf{A}^{t}_{l} \mathbf{k}_{l} = 
    \left[\begin{array}{rrr}
            -1&0&0\\
             1&0&0\\
             0&-1&0\\
             0&1&0\\
             0&0&-1\\
             0&0&1
          \end{array}\right]
    \left[\begin{array}{r} 
            1\\1\\1 
          \end{array}\right] 
          \frac{w}{6} = \frac{w}{6} 
          \left[\begin{array}{r}
                  -1\\
                  1\\
                  -1\\
                  1\\
                  -1\\
                  1 
                \end{array}\right]
$$
Then from equation (10.10),\\
然后从方程\eqref{eq:10-10}
$$
    \mathbf{Q}_{\mathbf{v}_d \mathbf{v}_d} = \mathbf{Q}_{d d} \mathbf{D}^{t} \mathbf{A}^{t}_{l} \mathbf{W}_{e_l} \mathbf{A}_{l} \mathbf{D} \mathbf{Q}_{d d} = \frac{1}{6} 
    \left[\begin{array}{rrrrrr} 
        1&-1&1&-1&1&-1\\
         &1&-1&1&-1&1\\
         & &1&-1&1&-1\\
         & & &1&-1&1\\
         & & & &1&-1\\
         & & & & &1 
    \end{array}\right]
$$
Both $v_d$  and $Q_{v_d v_d}$  are the same as was obtained previously.\\
$v_d$和$Q_{v_d v_d}$与之前获得的相同。

This verifies the relations derived in the present section.\\
这将验证本节中推导的关系。

Example 10.2.In this example the results are used from the adjustment in Example 7.5 of a quadrilateral by angles and directions in order to verify further the applicability of equations (10.9) and (10.10).\\
示例 10.2.在此示例中，使用四边形示例 7.5 中按角度和方向进行平差的结果，以进一步验证方程\eqref{eq:10-9}和\eqref{eq:10-10}的适用性。

Solution: Using equation \eqref{eq:10-9b}\\
解决方案：使用方程\eqref{eq:10-9b}
$$
    \underset{1 2, 1 2}{\mathbf{v}_{d}} = \underset{1 2, 1 2}{\mathbf{Q}_{d d}} \quad \underset{1 2, 8}{\mathbf{D}^{t}} \quad \underset{8,8}{\mathbf{Q}^{-1}_{l l}} \quad \underset{8,1}{\mathbf{v}_{l}}
$$
Where\\
$\mathbf{Q}_{d d} = \mathbf{I}_{1 2} \quad $and$ \quad \mathbf{D} = matrix \underset{8,1 2}{\mathbf{F}} $  in Example 7.5
$$
    \underset{8,8}{\mathbf{Q}^{-1}_{l l}} = \frac{1}{3} 
    \left[\begin{array}{rrrrrrrr}
            2&1&0&0&0&0&0&0\\
            1&2&0&0&0&0&0&0\\
            0&0&2&1&0&0&0&0\\
            0&0&1&2&0&0&0&0\\
            0&0&0&0&2&1&0&0\\
            0&0&0&0&1&2&0&0\\
            0&0&0&0&0&0&2&1\\
            0&0&0&0&0&0&1&2
          \end{array}\right]
$$

$$
    \mathbf{v}_{t} = \left[\begin{array}{rrrrrrrr}
        -1.0122 & 0.9822 & -0.3378 & 1.7137 & 1.7620 & -0.3722 & 1.1065 & -1.1022
    \end{array}\right]^t
$$

$$
    \mathbf{v}_{d} = \frac{1}{3}
    \left[\begin{array}{rrrrrrrr}
        -2 & -1 & 0  & 0  & 0 & 0  & 0  & 0  \\
        1  & -1 & 0  & 0  & 0 & 0  & 0  & 0  \\
        1  & 2  & 0  & 0  & 0 & 0  & 0  & 0  \\
        0  & 0  & -2 & -1 & 0 & 0  & 0  & 0  \\
        0  & 0  & 1  & 1  & 0 & 0  & 0  & 0  \\
        0  & 0  & 0  & 0  & 2 & -1 & 0  & 0  \\
        0  & 0  & 0  & 0  & 1 & -1 & 0  & 0  \\
        0  & 0  & 0  & 0  & 1 & 2  & 0  & 0  \\
        0  & 0  & 0  & 0  & 0 & 0  & -2 & -1 \\
        0  & 0  & 0  & 0  & 0 & 0  & 1  & -1 \\
        0  & 0  & 0  & 0  & 0 & 0  & 1  & 2
    \end{array}\right]
    \left[\begin{array}{r}
           -1.0122 \\
            0.9822 \\
           -0.3378 \\
            1.7137 \\
            1.7620 \\
           -0.3722 \\
            1.1065 \\
           -0.3660
    \end{array}\right] 
    = 
    \left[\begin{array}{r}
            0.3474\\
            -0.6648\\
            0.3174\\
            -0.3460\\
            -0.6838\\
            1.0299\\
            -1.0506\\
            0.7114\\
            0.3392\\
            -0.3703\\
            0.7362\\
            -0.3660
          \end{array}\right]
$$
which is exactly the same vector obtained from direct adjustment of girectiomns (see Example7.5).\\
这是从直接平差中获得的相同向量。（参见示例7.5）

With regard to equation (10.10), compute   as follows:\\
关于方程\eqref{eq:10-10}，计算$Q_{v_d v_d}$如下所示：
$$
    \mathbf{Q}_{v_d v_d} = \mathbf{Q}_{l l} \mathbf{D}^{t} \mathbf{A}^{t}_{l} \mathbf{W}_{e_l} \mathbf{A}_{l} \mathbf{D} \mathbf{Q}_{l l}=\underset{1 2, 8}{\mathbf{D}} \quad \underset{8,4}{\mathbf{A}^{t}_{l}} \quad \underset{4,4}{\mathbf{W}_{e_l}} \quad \underset{4,8}{\mathbf{A}_{l}} \quad \underset{8,1 2}{\mathbf{D}}
$$
Where
$$
    \underset{4,8}{\mathbf{A}_{t}} = \left[\begin{array}{llllllll}
        1      & 1       & 1      & 0       & 0      & 0       & 0      & 1       \\
        0      & 0       & 0      & 1       & 1      & 1       & 1      & 0       \\
        1      & 0       & 0      & 0       & 0      & 1       & 1      & 1       \\
        2.4716 & -3.3244 & 0.6457 & -2.8156 & 0.0602 & -0.6062 & 3.7025 & -0.1037
    \end{array}\right]
$$
$$
    \underset{4,4}{\mathbf{W}_{e_l}} = 
    \left[\begin{array}{rrrr}
        0.2773  & 0.1490  & -0.1765 & 0.0195  \\
                & 0.2712  & -0.1703 & 0.0171  \\
                &         & 0.3471  & -0.0367 \\
                &         &         & 0.0139
    \end{array}\right]
$$

The value of  $\mathbf{D}^{t} \mathbf{A}^{t}_{l}$ is first computed\\
首先计算 A 的值
$$
    \mathbf{D}^{t} \mathbf{A}^{t}_{l} =
    \left[\begin{array}{rrrr}
        -1& 0&-1&-2.4716\\
         0& 0& 1& 5.7960\\
         1& 0& 0&-3.3244\\
         1& 0& 0&-0.6457\\
         1&-1& 0& 3.4613\\
         0& 1& 0&-2.8156\\
         0& 1& 0&-0.0602\\
         0& 0&-1& 0.6644\\
         0& 1& 1&-0.6062\\
         0&-1&-1&-3.7026\\
        -1& 1& 0& 3.8062\\
         1& 0& 1&-0.1037    
    \end{array}\right]
$$


This is equal to $\mathbf{A}^{t}_{d}$ , and since $\mathbf{W}_{e_{d}} = \mathbf{W}_{e_{l}}$ , the computed $\mathbf{Q}_{v_{d} v_{d}}$  will be exactly the same as that computed from the adjustment of directions and need not be repeated.\\
这等于 $\mathbf{A}^{t}_{d}$ ，由于 $\mathbf{W}_{e_{d}} = \mathbf{W}_{e_{l}}$，计算 $\mathbf{W}_{e_{d}} = \mathbf{W}_{e_{l}}$ , 将和方向平差中计算的完全相同，无需重复。

Example 10.3. Refer to Figure 10.1 of the plane triangle and consider that the angle $A$ is measured twice ( $l_1$ and $l_2$ ), whereas angles $B$ and $C$ are each measured once $l_3$ and $l_4$ , respectively.\\
示例 10.3.参考平面三角形的图10.1，认为角度 $A$ 测量两次（$l_1$ 和 $l_2$），而角度 $B$ 和 $C$ 分别测量一次 $l_3$ 和 $l_4$。

The four measurements, which are uncorrelated and of equal precision, are\\
不相关和同等精度的四个测量，是
$$
    \begin{aligned}
        \boldsymbol{l}_{1} &= 60^\circ 00'\\
        \boldsymbol{l}_{2} &= 60^\circ 02'\\
        \boldsymbol{l}_{3} &= 60^\circ 00'\\
        \boldsymbol{l}_{4} &= 60^\circ 01'
    \end{aligned}
$$

It is required to adjust the triangle, first using all four observations and then using derived observations, and to check the consistency of both results.\\
对三角形平差首先使用所有四个观测值，然后使用派生观测值，并检查两个结果的一致性。

Solution: The model for this problem requires a minimum of two variables;\\
解决方案：此问题的模型至少需要两个变量

thus r = 2, and two conditions are needed as follows:\\
因此 $r=2$，需要以下两个条件：
$$
    \begin{aligned}
        \boldsymbol{l}_{1} - \boldsymbol{l}_{2} &= 0\\
        \boldsymbol{l}_{1} + \boldsymbol{l}_{3} + \boldsymbol{l}_{4} - \pi  &= 0
    \end{aligned}
$$
Or
$$
    \left[\begin{array}{rrrr}
        1&-1& 0& 0\\
        1& 0& 1& 1
    \end{array}\right]
    \mathbf{v}_{l} = 
    \left[\begin{array}{r}
        2'\\
        -1'
    \end{array}\right]
    \quad \text{with} \quad \mathbf{Q}_{l l} = \mathbf{I}_{4}
$$
Thus\\
因此
$$
    \mathbf{k}_{l} = 
    \left[\begin{array}{rr}
        2&1\\
        1&3
    \end{array}\right]^{-1}
    \left[\begin{array}{r}
        2\\
        -1
    \end{array}\right] = 
    \frac{1}{5} 
    \left[\begin{array}{r}
        7\\
        -4
    \end{array}\right]
$$

$$
    \mathbf{v}_{l} = \frac{1}{5} 
    \left[\begin{array}{rr}
        1&1  \\
       -1&0  \\
        0&1  \\
        0&1 
    \end{array}\right]
    \left[\begin{array}{r}
        7\\
        -4
    \end{array}\right] = \frac{1}{5} 
    \left[\begin{array}{r}
        3'\\
       -7'\\
       -4'\\
       -4' 
    \end{array}\right] 
    \quad \mathbf{Q}_{ v_{l} v_{l}} = \frac{1}{5} 
    \left[\begin{array}{rrrr}
        3&-2& 1& 1\\
       -2& 3& 1& 1\\
        1& 1& 2& 2\\
        1& 1& 2& 2 
    \end{array}\right]
$$
It is possible to first derive one “observation” from   and  . Since both are of equal precision, then the derived observations are\\
有可能首先从$\boldsymbol{l}_{1}$和$\boldsymbol{l}_{2}$中得出一个"观察值"。由于两者都是同等精度的，因此派生的观测值为
$$
    \left[\begin{array}{l}
        \mathbf{m}_{1}\\
        \mathbf{m}_{2}\\
        \mathbf{m}_{3}
    \end{array}\right] = 
    \left[\begin{array}{rrrr}
        \frac{1}{2}&\frac{1}{2}&0&0\\
                  0&          0&1&0\\
                  0&          0&0&1
    \end{array}\right]
    \left[\begin{array}{r}
        \boldsymbol{l}_{1}\\
        \boldsymbol{l}_{2}\\
        \boldsymbol{l}_{3}\\
        \boldsymbol{l}_{4}
    \end{array}\right] 
    \quad \text{or} \quad
    \mathbf{m} = \mathbf{D} \boldsymbol{l}
$$
where $ \mathbf{m}_1 $  is the arithmetic mean of $ \boldsymbol{l}_1 $  and $ \boldsymbol{l}_2 $ .\\
其中 $ \mathbf{m}_1 $ 是 $ \boldsymbol{l}_1 $ 和 $ \boldsymbol{l}_2 $ 的算术平均值。

With these three observations, we may now adjust the triangle.\\
有了这三个观测值，我们现在可以对三角形平差。

However, with only three observations, we would have only one redundancy.\\
但是，只有三个观察值，我们就只有一个冗余数据。

This means that in the process of obtaining the derived observations we also have reduced the redundancy from two to one. \\
这意味着，在获取派生观测值的过程中，我们也将冗余数据从 2 减少到 1。

This makes it no longer a technique using derived observations, and thus a treatment using adjustment in steps is warranted instead.\\
这使得它不再是一种使用派生观测的技术，因此有必要使用分步平差进行改正。

First adjust $\boldsymbol{l}_1$ and $\boldsymbol{l}_2$ , and in the second part use the adjusted values as derived observations.\\
首先对 $\boldsymbol{l}_1$ 和$\boldsymbol{l}_2$平差，在第二部分使用平差值作为派生观测值。

This is an adjustment in steps, as will be explained in the next section.\\
这是分步平差，如下一节所述。

Adjusting $\boldsymbol{l}_1$, $\boldsymbol{l}_2$ first we get
对$\boldsymbol{l}_1$, $\boldsymbol{l}_2$的平差中首先我们得到
$$
    v_1 = + 1' \quad v_2 = - 1'\\
$$

$$
    \hat{\boldsymbol{l}}_1 = 60^\circ 01' \quad \hat{\boldsymbol{l}}_2 = 60^\circ 01' 
$$
And
$$
    \hat{\mathbf{Q}} = 
    \left[\begin{array}{rr}
        \frac{1}{2}&\frac{1}{2}\\
        \frac{1}{2}&\frac{1}{2}
    \end{array}\right]
$$
Now we work with the “derived” observations m noting that we should use   and   instead of   and  .\\
现在，我们使用"派生"的观测值，指出我们应该使用$\hat{\boldsymbol{l}}_1$和$\hat{\boldsymbol{l}}_2$，而不是$\boldsymbol{l}_1$和$\boldsymbol{l}_2$。

Hence\\
因此
$$
    \underset{3,3}{\mathbf{Q}_{m m}} = 
    \left[\begin{array}{rrrr}
        \frac{1}{2}&\frac{1}{2}&0&0\\
                  0&          0&1&0\\
                  0&          0&0&1
    \end{array}\right]
    \left[\begin{array}{rrrr}
        \frac{1}{2}&\frac{1}{2}&0&0\\
        \frac{1}{2}&\frac{1}{2}&0&0\\
                  0&          0&1&0\\
                  0&          0&0&1
    \end{array}\right]
    \left[\begin{array}{rrr}
        \frac{1}{2}&0&0\\
        \frac{1}{2}&0&0\\
                  0&1&0\\
                  0&0&1
    \end{array}\right] = 
    \left[\begin{array}{rrr}
        \frac{1}{2}&0&0\\
                  0&1&0\\
                  0&0&1
    \end{array}\right]
$$
And the condition  $\mathbf{A}_{m} \mathbf{V}_m = \mathbf{f}_m$ is\\
条件$\mathbf{A}_{m} \mathbf{V}_m = \mathbf{f}_m$是
$$
    \left[\begin{array}{lll}
        1&1&1
    \end{array}\right] 
    \mathbf{V}_{m} = 180^\circ - 60^\circ 00' - 60^\circ 01' = -2'
$$

$$
    \mathbf{k}_{m} = \frac{2}{5} (-2) = -\frac{4'}{5}
$$

$$
    \mathbf{v}_{m} = 
    \left[\begin{array}{rrr}
        \frac{1}{2}&0&0\\
                  0&1&0\\
                  0&0&1
    \end{array}\right]
    \left[\begin{array}{r}
        1\\
        1\\
        1
    \end{array}\right] (-\frac{4}{5}) = \frac{1}{5} 
    \left[\begin{array}{r}
        -2\\
        -4\\
        -4
    \end{array}\right]
$$
Now, in order to compute $\mathbf{v}_{l'}$  apply equation (10.9a)\\
现在，为了计算$\mathbf{v}_{l}$应用方程\eqref{eq:10-9a}
$$
    \mathbf{v}_{l'} = \mathbf{Q}_{l' l'} \mathbf{D}^{t} \mathbf{A}^{t}_{m} \mathbf{k}_{m} = 
    \left[\begin{array}{rrrr}
        \frac{1}{2}&\frac{1}{2}&0&0\\
        \frac{1}{2}&\frac{1}{2}&0&0\\
                  0&          0&1&0\\
                  0&          0&0&1
    \end{array}\right]
    \left[\begin{array}{rrr}
        \frac{1}{2}&0&0\\
        \frac{1}{2}&0&0\\
                  0&1&0\\
                  0&0&1
    \end{array}\right]
    \left[\begin{array}{r}
        1\\
        1\\
        1
    \end{array}\right] (-\frac{4}{5}) = \frac{1}{5}
    \left[
        \begin{array}{r}
            -2\\
            -2\\
            -4\\
            -4
        \end{array}
    \right]
$$

Note the first two correction $-\frac{2}{5}$  and $-\frac{2}{5}$  are to be applied to $\hat{\boldsymbol{l}}_{1}$  and $\hat{\boldsymbol{l}}_{2}$  ,respectively, after the first step of adjustment ,and not to the original observations $\boldsymbol{l}_{1}$  and $\boldsymbol{l}_{2}$  .\\
请注意， 第一步平差之后，前两个修正 $-\frac{2}{5}$ 和 $-\frac{2}{5}$ 将分别应用于 $\hat{\boldsymbol{l}}_{1}$ 和 $\hat{\boldsymbol{l}}_{2}$，而不是原始观测值 $\boldsymbol{l}_{2}$ 和 $\boldsymbol{l}_{2}$。 

In order to get the total residuals,add the contribution from the step to that from the adjustment eith reduced observations,or\\
为了获得总残差，将步骤中的累积添加到平差中减小的观测值，或
$$
    \mathbf{v}_{l} = 
    \left[\begin{array}{r}
        1'\\
        -1'\\
        0\\
        0
    \end{array}\right] + \frac{1}{5} 
    \left[\begin{array}{r}
        -2'\\
        -2'\\
        -4'\\
        -4'
    \end{array}\right] = 
    \left[\begin{array}{r}
        3'\\
        -7'\\
        -4'\\
        -4'\\
    \end{array}\right]
$$

Which is identical to that obtained directly from the adjustment of four observations. \\
与直接从四个观测值的平差中获得的相同。

Next, we should evaluate and check the cofactor matrices of the estimated observations.\\
接下来，我们应评估和检查估计观测值的余因数矩阵。

(Check the derivations in the next section to ascertain this point .)\\
（检查下一节中的派生，以确定此点。）

Thus first compute\\
因此，首先计算
$$
    \underset{4,4}{\mathbf{Q}_{l l}} \quad \text{from} \quad \mathbf{Q}_{v_{l} v_{l}}
$$
evaluated in the first part\\
在第一部分评估
$$
    \mathbf{Q}_{l l} = \mathbf{I} - \frac{1}{5} 
    \left[\begin{array}{rrrr}
        3&-2& 1& 1\\
           3& 1& 1\\
              2& 2\\
                 2
    \end{array}\right] = \frac{1}{5} 
    \left[\begin{array}{rrrr}
        2&2&-1&-1\\
          2&-1&-1\\
             3&-2\\
                3
    \end{array}\right]
$$
Then compute $\mathbf{Q}_{v_{l'} v_{l'}}$  ,from the second part (with derived observations) using equation
(10.10), or\\
然后计算 $\mathbf{Q}_{v_{l'} v_{l'}}$ ，从第二部分（与派生观测值）使用方程\eqref{eq:10-10}，或
$$
    \mathbf{Q}_{v_{l'} v_{l'}} = \mathbf{Q}_{l' l'} \mathbf{D}^{t} \mathbf{A}^{t}_{m} \mathbf{W}_{e_{m}} \mathbf{A}_{m} \mathbf{D} \mathbf{Q}_{l' l'}
$$

$$
    \mathbf{Q}_{v_{l'} v_{l'}} = \frac{1}{10} 
    \left[\begin{array}{rrrr}
        1&1&2&2\\
          1&2&2\\
            4&4\\
              4
    \end{array}\right]
$$
Finally,\\
最后
$$
    \mathbf{Q}_{l l} = \frac{1}{5} 
    \left[\begin{array}{rrrr}
        2&2&-1&-1\\
          2&-1&-1\\
             3&-2\\
                3
    \end{array}\right]
$$
This value is identical to that computed by the other procesure .\\
这个结果与其他结果计算的值相同。

A last and important remark: It is worth nothing that although it may not have been readily evident, it was possible to propagate backward from a smaller set of variables(for example,angles)to a larger set of variables(directions).\\
最后一句重要的话：虽然它可能不显而易见，但有可能从较小的变量集（例如角度）向后传播到一组较大的变量（方向），这一点毫无意义。

\section{分步平差}

We begin adjustment in steps by having a total set of(r+s) condition equtions\\
我们通过一组($r + s$)条件等价开始分步平差
$$
    \underset{(r + s, n)}{\mathbf{A}} \left( \boldsymbol{l} + \underset{n,1}{v} \right) = \underset{(r + s), 1}{d}
$$
Amongst the observations  $\boldsymbol{l}_{n,1}$ ,the cofactor matrix of which is $\mathbf{Q}$  .\\
在观测值 $\boldsymbol{l}_{n,1}$ 中，其余因数矩阵为$\mathbf{Q}$。

The solution for this total system is given by equations(7.4),(7.5),and(7.7)in Section10.2.\\
整个系统的解由第 10.2 节中的方程\eqref{eq:7-4}、\eqref{eq:7-5}和\eqref{eq:7-7}给出。
Due to the manner in which the conditions arise,the total system of conditions is partitioned into two systems,the first being\\
由于条件的产生方式，整个条件系统被划分为两个系统，第一是
\begin{equation}
    \underset{r,n}{\mathbf{A}_{1}} \left( \boldsymbol{l} + \underset{n,1}{\mathbf{v}_{1}} \right) = \underset{r,1}{\mathbf{d}_{1}} \label{eq:10-11}
\end{equation}
Using the equations just referred to,we get   ,  ,and  as the solution for the first system in the first step .\\
使用刚才提到的方程，我们得到$v_{1}$，$\mathbf{Q}_{v_1 v_1}$和$\mathbf{Q}_{2} = \mathbf{Q} - \mathbf{Q}_{v_1 v_1}$作为第一个系统第一步的解。

Next,the second system of conditions is\\
接下来，第二个条件系统是
\begin{equation}
    \underset{s,n}{\mathbf{A}_{2}} \left( \boldsymbol{l} + \underset{n,1}{\mathbf{v}_{1}} + \mathbf{v}_{2} \right) = \underset{s,1}{\mathbf{d}_{2}} \label{eq:10-12}
\end{equation}
For which the cofactor matrix is now  $\mathbf{Q}_{2}$ .\\
余因数矩阵现在是$\mathbf{Q}_{2}$。

Again using the same equations,solve for $\mathbf{v}_{2}$  and $\mathbf{Q}_{v_2 v_2}$  and compute\\
再次使用相同的方程，求解$\mathbf{v}_{2}$和$\mathbf{Q}_{v_2 v_2}$并计算
\begin{equation}
    \bar{\mathbf{Q}}_{l l} = \mathbf{Q} - \mathbf{Q}_{v_2 v_2} \label{eq:10-13}
\end{equation}
The answer obtained from the tatal system of conditions may be obtained
From the adjustment in steps as follows:\\
从条件系统获得的答案也许可以从分步平差中获得，步骤如下：
\begin{subequations}
    \begin{align}
        \mathbf{v}       &= \mathbf{v}_{1} + \mathbf{v}_{2} \label{eq:10-14a} \\
        \mathbf{Q}_{l l} &= \bar{\mathbf{Q}}_{l l} \label{eq:10-14b} \\
        \phi             &= \phi_{1} + \phi_{2} \quad \text{( the quadratic forms)} \label{eq:10-14c}
        \end{align} \label{eq:10-14}
\end{subequations}

10.3.1 Proof that $\mathbf{v} = \mathbf{v}_{1} + \mathbf{v}_{2}$\\
10.3.1 证明$\mathbf{v} = \mathbf{v}_{1} + \mathbf{v}_{2}$

The total system may be partitioned to\\
整个系统可以分区为
$$
    \left[\begin{array}{r}
        \mathbf{A}_{1}\\
        \mathbf{A}_{2}
    \end{array}\right]
    \left( \boldsymbol{l} + \mathbf{v} \right) = 
    \left[\begin{array}{r}
        \mathbf{d}_{1}\\
        \mathbf{d}_{2}
    \end{array}\right]
$$
Or
\begin{subequations}
    \begin{align}
        \mathbf{A}_{1} \mathbf{v} = \mathbf{d}_{1} - \mathbf{A}_{1} \boldsymbol{l} = \mathbf{f}_{1} \label{eq:10-15a}\\
        \mathbf{A}_{2} \mathbf{v} = \mathbf{d}_{2} - \mathbf{A}_{2} \boldsymbol{l} = \mathbf{f}_{2} \label{eq:10-15b}
    \end{align}\label{eq:10-15}
\end{subequations}
Now with regard to the adjustment in steps,from the first step:\\
现在关于分步平差，从第一步：
\begin{subequations}
    \begin{align}
        \mathbf{A}_{1} \mathbf{v}_{1} &= \mathbf{d}_{1} - \mathbf{A}_{1} \boldsymbol{l} = \mathbf{f}_{1} \label{eq:10-16a}\\
        \mathbf{Q}_{v_1 v_1}          &= \mathbf{Q} \mathbf{A}^{t}_{1} \left( \mathbf{A}_{1} \mathbf{Q} \mathbf{A}^{t}_{1} \right)^{-1} \mathbf{A}_{1} \mathbf{Q} \label{eq:10-1bb} \\
        \mathbf{Q}_{2}                &= \mathbf{Q} - \mathbf{Q} \mathbf{A}^{t}_{1} \left( \mathbf{A}_{1} \mathbf{Q} \mathbf{A}^{t}_{1} \right)^{-1} \mathbf{A}_{1} \mathbf{Q} \label{eq:10-1bc}
    \end{align} \label{eq:10-16}
\end{subequations}
For the second step\\
第二步
\begin{equation}
        \mathbf{A}_{2} \left( \boldsymbol{l} + \mathbf{v}_{1} + \mathbf{v}_{2} \right) = \mathbf{d}_{2}\\
        \mathbf{A}_{2} \left( \mathbf{v}_{1} + \mathbf{v}_{2} \right) = \mathbf{d}_{2} - \mathbf{A}_{2} \boldsymbol{l}
\end{equation} \label{eq:10-17}

From equations (10.7) and (10.15) it follows that\\
从方程\eqref{eq:10-7}和\eqref{eq:10-15}， 它遵循
$$
    \mathbf{v} = \mathbf{v}_{1} + \mathbf{v}_{2}
$$

Provided that the consistency between equations (10.15a) and (10.16b) is ascertained.\\
前提是确定方程\eqref{eq:10-15a}和\eqref{eq:10-16a}之间的一致性。

For such a consistency to stand we must prove that\\
要保持这种一致性，我们必须证明
$$
    \mathbf{A}_{1} \mathbf{v}_{2} = 0
$$
Solving equation (10.17) for $\mathbf{v}_{2}$\\
$\mathbf{v}_{2}$求解方程\eqref{eq:10-17}
$$
    \begin{aligned}
        \mathbf{v}_{2} & = \mathbf{Q}_{2} \mathbf{A}^{t}_{2} \left( \mathbf{A}_{2} \mathbf{Q}_{2} \mathbf{A}^{t}_{2} \right)^{-1} \left( \mathbf{d}_{2} - \mathbf{A}_{2} \mathbf{I} - \mathbf{A}_{2} \mathbf{v}_{1} \right)\\
                       & = \mathbf{Q}_{2} \mathbf{A}^{t}_{2} \left( \mathbf{A}_{2} \mathbf{Q}_{2} \mathbf{A}^{t}_{2} \right)^{-1} \mathbf{f'}_{2}
    \end{aligned}
$$
And using equation (10.16c)\\
并用方程\eqref{eq:10-16c}
$$
    \mathbf{v}_{2} = \left[ \mathbf{Q} - \mathbf{Q} \mathbf{A}^{t}_{1} \left( \mathbf{A}_{1} \mathbf{Q} \mathbf{A}^{t}_{1}\right)^{-1} \mathbf{A}_{1} \mathbf{Q} \right] \mathbf{A}^{t}_{2} \left(\mathbf{A}_{2} \mathbf{Q}_{2} \mathbf{A}^{t}_{2}\right)^{-1} \mathbf{f'}_{2}
$$
Now premultiplying by A1\\
现在由$\mathbf{A}_{1}$乘
$$
    \begin{aligned}
        \mathbf{A}_{1} \mathbf{v}_{2} &= \left[ \mathbf{A}_{1} \mathbf{Q} - \left( \mathbf{A}_{1} \mathbf{Q} \mathbf{A}^{t}_{1}\right) \left( \mathbf{A}_{1} \mathbf{Q} \mathbf{A}^{t}_{1}\right)^{-1} \mathbf{A}_{1} \mathbf{Q} \right] \mathbf{A}^{t}_{2} \left( \mathbf{A}_{2} \mathbf{Q}_{2} \mathbf{A}^{t}_{2} \right)^{-1} \mathbf{f'}_{2}\\
                                      &= 0
    \end{aligned}
$$
Thus proving the validity of equation (10.14a).\\
从而证明方程\eqref{eq:10-14a}的有效性。

10.3.2. Proof that $\mathbf{Q}_{l l} = \bar{\mathbf{Q}}_{l l}$  we first evaluate $\bar{\mathbf{Q}}_{l l}$  from the total system after it is partitioned.\\
10.3.2. 证明$\mathbf{Q}_{l l} = \bar{\mathbf{Q}}_{l l}$   我们首先在总系统中对$\bar{\mathbf{Q}}_{l l}$进行分区后对其进行评估。
$$
    \mathbf{Q} = \mathbf{Q} - \mathbf{Q} \mathbf{A}^{t} \left( \mathbf{A} \mathbf{Q} \mathbf{A}^{t} \right)^{-1} \mathbf{A} \mathbf{Q}
$$
For which we expand\\
为此我们扩展
$$
    \begin{aligned}
        \left( \mathbf{A} \mathbf{Q} \mathbf{A}^{t} \right)^{-1} 
        & = 
        \left\{
            \left[\begin{array}{r}
                \mathbf{A}_{1}\\
                \mathbf{A}_{2}
            \end{array}\right]
            \mathbf{Q}
            \left[\begin{array}{rr}
                \mathbf{A}^{t}_{1}&\mathbf{A}^{t}_{2}
            \end{array}\right]
        \right\}^{-1} = 
        \left[\begin{array}{rr}
            \mathbf{A}_{1} \mathbf{Q} \mathbf{A}^{t}_{1}&\mathbf{A}_{1} \mathbf{Q} \mathbf{A}^{t}_{2}\\
            \mathbf{A}_{2} \mathbf{Q} \mathbf{A}^{t}_{1}&\mathbf{A}_{2} \mathbf{Q} \mathbf{A}^{t}_{2}
        \end{array}\right]^{-1}\\
        & =
        \left[\begin{array}{rr}
                \dot{\mathbf{M}}&\bar{\mathbf{M}}\\
            \bar{\mathbf{M}}^{t}&\ddot{\mathbf{M}}
        \end{array}\right]^{-1} = 
        \left[\begin{array}{rr}
                \mathbf{E}&\mathbf{G}\\
            \mathbf{G}^{t}&\mathbf{H}
        \end{array}\right]
    \end{aligned}  
$$

Where\\
\begin{subequations}
    \begin{align}
        \mathbf{H} &= \left( \ddot{\mathbf{M}} - \bar{\mathbf{M}}^{t} \dot{\mathbf{M}}^{-1} \bar{\mathbf{M}} \right)^{-1}\label{eq:10-18a}\\
        \mathbf{G} &= -\dot{\mathbf{M}} ^{-1} \bar{\mathbf{M}} \mathbf{H}\label{eq:10-18b}\\
        \mathbf{E} &= \dot{\mathbf{M}} ^{-1} - \dot{\mathbf{M}} ^{-1} \bar{\mathbf{M}} \mathbf{G}^{t}\label{eq:10-18c}
    \end{align}
\end{subequations}
Thus\\
因此
$$
    \mathbf{Q}_{l l} = \mathbf{Q} - \mathbf{Q} \left[ \mathbf{A}^{t}_{1} \mathbf{A}^{t}_{2}\right]
    \left[\begin{array}{rr}
            \mathbf{E}&\mathbf{G}\\
        \mathbf{G}^{t}&\mathbf{H}
    \end{array}\right]
    \left[\begin{array}{r}
        \mathbf{A}_{1}\\
        \mathbf{A}_{2}
    \end{array}\right] \mathbf{Q}
$$
Or
$$
    \begin{aligned}
        \mathbf{Q}_{l l} = & \mathbf{Q} - \mathbf{Q} \mathbf{A}^{t}_{1} \mathbf{E} \mathbf{A}_{1} \mathbf{Q} - \mathbf{Q} \mathbf{A}^{t}_{2} \mathbf{G}^{t} \mathbf{A}_{1} \mathbf{Q} - \mathbf{Q} \mathbf{A}^{t}_{1} \mathbf{G} \mathbf{A}_{2} \mathbf{Q} - \mathbf{Q} \mathbf{A}^{t}_{2} \mathbf{H} \mathbf{A}_{2} \mathbf{Q}\\
                         = & \mathbf{Q} - \mathbf{Q} \mathbf{A}^{t}_{1} \left( \dot{\mathbf{M}}^{-1} - \dot{\mathbf{M}}^{-1} \bar{\mathbf{M}} \mathbf{G}^{t} \right) \mathbf{A}_{1} \mathbf{Q} - \mathbf{Q} \mathbf{A}^{t}_{2} \mathbf{G}^{t} \mathbf{A}^{1} \mathbf{Q} \\
                           & -\mathbf{Q} \mathbf{A}^{t}_{1} \mathbf{G} \mathbf{A}_{2} \mathbf{Q} - \mathbf{Q} \mathbf{A}^{t}_{2} \mathbf{H} \mathbf{A}_{2} \mathbf{Q}
    \end{aligned}
$$
Or
\begin{equation}
    \begin{aligned}
        \mathbf{Q}_{l l} = & \mathbf{Q} - \underset{(2)}{\left( \mathbf{Q} \mathbf{A}^{t}_{1} \mathbf{M}^{-1} \mathbf{A}_{1} \mathbf{Q} \right)} + \underset{(3)}{\left( \mathbf{Q} \mathbf{A}^{t}_{1} \dot{\mathbf{M}}^{-1} \bar{\mathbf{M}} \mathbf{G}^{t} \mathbf{A}_{1} \mathbf{Q} \right)}\\
                           & -\underset{(4)}{\left( \mathbf{Q} \mathbf{A}^{t}_{2} \mathbf{G}^{t} \mathbf{A}_{1} \mathbf{Q} \right)} - \underset{(5)}{\left( \mathbf{Q} \mathbf{A}^{t}_{1} \mathbf{G} \mathbf{A}_{2} \mathbf{Q} \right)}\\
                           & -\underset{(6)}{\left( \mathbf{Q} \mathbf{A}^{t}_{2} \mathbf{H} \mathbf{A}_{2} \mathbf{Q} \right)}
    \end{aligned} \label{eq:10-19}
\end{equation}
Then we also evaluate $\bar{\mathbf{Q}}_{l l}$  as follows:\\
然后，我们还对$\bar{\mathbf{Q}}_{l l}$进行如下评估：
\begin{equation}
    \begin{aligned}
        \bar{\mathbf{Q}}_{l l} = & \mathbf{Q}_{2} - \mathbf{Q}_{2} \mathbf{A}^{t}_{2} \left( \mathbf{A}_{2} \mathbf{Q}_{2} \mathbf{A}^{t}_{2} \right)^{-1} \mathbf{A}^{t}_{2} \mathbf{Q}_{2}\\
                               = & \underset{(1)}{\mathbf{Q}} - \underset{(2)}{\mathbf{Q} \mathbf{A}^{t}_{1} \dot{\mathbf{M}}^{-1} \mathbf{A}^{t}_{1} \mathbf{Q}} \\
                                 & - \left( \mathbf{Q} - \mathbf{Q} \mathbf{A}^{t}_{1} \dot{\mathbf{M}}^{-1} \mathbf{A}^{t}_{1} \mathbf{Q} \right) \mathbf{A}_{2} \left( \mathbf{A}_{2} \mathbf{Q}_{2} \mathbf{A}^{t}_{2} \right)^{-1} \mathbf{A}_{2} \left( \mathbf{Q} - \mathbf{Q} \mathbf{A}^{t}_{1} \dot{\mathbf{M}}^{-1} \mathbf{A}^{t}_{1} \mathbf{Q} \right)
    \end{aligned} \label{eq:10-20}
\end{equation}
We evaluate the following term first:\\
我们首先评估以下术语：
\begin{equation}
    \begin{aligned}
        \left( \mathbf{A}_{2} \mathbf{Q}_{2} \mathbf{A}^{t}_{2} \right)^{-1} &= \left[ \mathbf{A}_{2} \left( \mathbf{Q} - \mathbf{Q} \mathbf{A}^{t}_{1} \dot{\mathbf{M}}^{-1} \mathbf{A}^{t}_{1} \mathbf{Q} \right) \mathbf{A}^{t}_{2} \right]^{-1}\\
                                                                             &= \left[\left( \mathbf{A}_{2} \mathbf{Q} \mathbf{A}^{t}_{2} \right) - \left( \mathbf{A}_{2} \mathbf{Q} \mathbf{A}^{t}_{1}\right) \mathbf{M}^{-1} \left( \mathbf{A}_{1} \mathbf{Q} \mathbf{A}^{t}_{2} \right) \right]^{-1}\\
                                                                             &= \left( \ddot{\mathbf{M}} - \bar{\mathbf{M}}^{t} \dot{\mathbf{M}}^{-1} \bar{\mathbf{M}} \right)^{-1}\\
                                                                             &= \mathbf{H}
    \end{aligned} \label{eq:10-21}
\end{equation}
Since the first terms are the same in both equation (10.19) and equation (10.20), we consider he last term in equation (10.20), which is denoted by trussing the just derived result.\\
由于第一个术语在方程\eqref{eq:10-19}和方程\eqref{eq:10-20}中是相同的，因此我们认为他在方程\eqref{eq:10-20}中的最后一个术语，该方程由$\mathbf{t}$使用刚刚推出的结果表示。
$$
    \begin{aligned}
        \mathbf{t} = & -\mathbf{Q} \mathbf{A}^{t}_{2} \mathbf{H} \mathbf{A}_{2} \mathbf{Q} + \mathbf{Q} \mathbf{A}^{t}_{2} \mathbf{H} \mathbf{A}_{2} \mathbf{Q} \mathbf{A}^{t}_{1} \dot{\mathbf{M}}^{-1} \mathbf{A}_{1} \mathbf{Q} + \mathbf{Q} \mathbf{A}^{t}_{1} \dot{\mathbf{M}}^{-1} \mathbf{A}_{1} \mathbf{Q} \mathbf{A}^{t}_{2}\\
                      & \times \mathbf{H} \mathbf{A}_{2} \mathbf{Q} - \mathbf{Q} \mathbf{A}^{t}_{1} \dot{\mathbf{M}}^{-1} \mathbf{A}_{1} \mathbf{Q} \mathbf{A}^{t}_{2} \mathbf{H} \mathbf{A}_{2} \mathbf{Q} \mathbf{A}^{t}_{1} \dot{\mathbf{M}}^{-1} \mathbf{A}_{1} \mathbf{Q}
    \end{aligned}
$$
Dropping the first term since it is the same as the sixth term of equation (10.19), we continue with
删除第一个术语，因为它与方程的第六个术语\eqref{eq:10-19}相同，我们继续
\begin{equation}
    \begin{aligned}
        \mathbf{t} = & \mathbf{Q} \mathbf{A}^{t}_{2} \mathbf{H} \bar{\mathbf{M}}^{t} \dot{\mathbf{M}}^{-1} \mathbf{A}_{1} \mathbf{Q} + \mathbf{Q} \mathbf{A}^{t}_{1} \dot{\mathbf{M}}^{-1} \bar{\mathbf{M}} \mathbf{H} \mathbf{A}_{2} \mathbf{Q}\\
                     & - \mathbf{Q} \mathbf{A}^{t}_{1} \dot{\mathbf{M}}^{-1} \bar{\mathbf{M}} \mathbf{H} \bar{\mathbf{M}}^{t} \dot{\mathbf{M}}^{-1} \mathbf{A}_{1} \mathbf{Q}\\
                   = & - \underset{(4)}{\mathbf{Q} \mathbf{A}^{t}_{2} \mathbf{G}^{t} \mathbf{A}_{1} \mathbf{Q}} - \underset{(5)}{\mathbf{Q} \mathbf{A}^{t}_{1} \mathbf{G} \mathbf{A}_{2} \mathbf{Q}} + \underset{(3)}{\mathbf{Q} \mathbf{A}_{1} \dot{\mathbf{M}}^{-1} \bar{\mathbf{M}} \mathbf{G}^{t} \mathbf{A}_{1} \mathbf{Q}} 
    \end{aligned} \label{eq:10-22}
\end{equation}
The first term of equation (10.21) is the same as the fourth term in equation (10.119), the second is the same as the fifth, and the last term is the same as the third, respectively. \\
方程的第一个术语\eqref{eq:10-21}与方程中的第四个术语\eqref{eq:10-19}相同，第二个术语与第五个术语相同，最后一个术语分别与第三个术语相同。

This proves that both   and   can be expanded to six terms that are identical. Hence
这证明$\mathbf{Q}_{l l}$和$\bar{\mathbf{Q}}_{l l}$都可以扩展到六个相同的术语。因此
$$
    \mathbf{Q}_{l l} = \bar{\mathbf{Q}}_{l l}
$$
10.3.3. Proof that$\phi = \phi_{1} + \phi_{2}$ In order to compute an posteriori estimate of the reference variance, the quadratic form $\phi = \mathbf{v}^{t} \mathbf{W} \mathbf{V}$ must be computed.\\
10.3.3 证明$\phi = \phi_{1} + \phi_{2}$ 为了计算参考方差的后向估计值，必须计算二次形式$\phi = \mathbf{v}^{t} \mathbf{W} \mathbf{V}$。

For the computation of $\phi$ we use equation (7.6) or \\
对于 $\phi$ 的计算，我们使用方程 （7.6） 或
$$
    \phi = \mathbf{k}^{t} \mathbf{f} = \mathbf{f}^{t} \left( \mathbf{A} \mathbf{Q} \mathbf{A}^{t} \right)^{-1} \mathbf{f}
$$
From equation (10.15b) and (10.17) we can readily write\\
从方程\eqref{eq:10-15b}和\eqref{eq:10-17}， 我们可以很容易地写出
\begin{equation}
    \mathbf{f'}_{2} = \left( \mathbf{d}_{2} - \mathbf{A}_{2} \boldsymbol{l} \right) - \mathbf{A}_{2} \mathbf{v}_{1} = \mathbf{f}_{2} - \mathbf{A}_{2} \mathbf{v}_{1} \label{eq:10-23}
\end{equation} 
And we also recall that\\
我们还记得
\begin{equation}
    \mathbf{v}_{1} = \mathbf{Q} \mathbf{A}^{t}_{1} \left( \mathbf{A}_{1} \mathbf{Q} \mathbf{A}^{t}_{1} \right)^{-1} \mathbf{f}_{1} = \mathbf{Q} \mathbf{A}^{t}_{1} \dot{\mathbf{M}}^{-1} \mathbf{f}_{1}\label{eq:10-24}
\end{equation}
\begin{equation}
    \begin{aligned}
        \phi &= \mathbf{f}^{t} 
        \left[\begin{array}{rr}
            \dot{\mathbf{M}}&\bar{\mathbf{M}}\\
            \bar{\mathbf{M}}^{t}&\ddot{\mathbf{M}}
        \end{array}\right]^{-1} \mathbf{f} = 
        \left[\begin{array}{rr}
            \mathbf{f}^{t}_{1}&\mathbf{f}^{t}_{2}
        \end{array}\right]
        \left[\begin{array}{rr}
            \mathbf{E}&\mathbf{G}\\
            \mathbf{G}^{t}&\mathbf{H}
        \end{array}\right]
        \left[\begin{array}{r}
            \mathbf{f}_{1}\\
            \mathbf{f}_{2}
        \end{array}\right]\\
            &= \left( \underset{(1)}{\mathbf{f}^{t}_{1} \dot{\mathbf{M}}^{-1} \mathbf{f}_{1}} - \underset{(2)}{\mathbf{f}^{t}_{1} \dot{\mathbf{M}}^{-1} \bar{\mathbf{M}} \mathbf{G}^{t} \mathbf{f}_{1}} \right) + \underset{(3)}{2 \mathbf{f}^{t}_{1} \mathbf{G} \mathbf{f}_{2}} + \underset{(4)}{\mathbf{f}^{t}_{2} \mathbf{H} \mathbf{f}_{2}} 
    \end{aligned}\label{eq:10-25}
\end{equation}
And
\begin{equation}
    \phi_{1} = \mathbf{f}^{t}_{1} \left( \mathbf{A}_{1} \mathbf{Q} \mathbf{A}^{t}_{1} \right)^{-1} \mathbf{f}_{1} = \mathbf{f}^{t}_{1} \dot{\mathbf{M}}^{-1} \mathbf{f}_{1}\label{eq:10-26}
\end{equation}
Finally,\\
最后
\begin{equation}
    \phi_{2} = \mathbf{f'}^{t}_{2} \left( \mathbf{A}_{2} \mathbf{Q}_{2} \mathbf{A}^{t}_{2} \right)^{-1} \mathbf{f'}_{2}\label{eq:10-27}
\end{equation}
Which according to equations (10.21), (10.23), and (10.24) becomes\\
根据方程\eqref{eq:10-21}，\eqref{eq:10-23}和\eqref{eq:10-24}成为
\begin{equation}
    \begin{aligned}
        \phi_{2} = & \mathbf{f'}^{t}_{2} \mathbf{H} \mathbf{f'}_{2}\\
                 = & \left( \mathbf{f}^{t}_{2} - \mathbf{v}^{t}_{1} \mathbf{A}^{t}_{2} \right) \mathbf{H} \left( \mathbf{f}_{2} - \mathbf{A}_{2} \mathbf{v}_{1} \right)\\
                 = & \mathbf{f}^{t}_{2} \mathbf{H} \mathbf{f}_{2} - 2\mathbf{v}^{t}_{1} \mathbf{A}^{t}_{2} \mathbf{H} \mathbf{f}_{2} + \mathbf{v}^{t}_{1} \mathbf{A}^{t}_{2} \mathbf{H} \mathbf{A}_{2} \mathbf{v}_{1}\\
                 = & \mathbf{f}^{t}_{2} \mathbf{H} \mathbf{f}_{2} - 2\mathbf{f}^{t}_{1} \dot{\mathbf{M}}^{-1}  \left( \mathbf{A}_{1} \mathbf{Q} \mathbf{A}^{t}_{2} \right) \mathbf{H} \mathbf{f}_{2}\\
                   & + \mathbf{f}^{t}_{1} \dot{\mathbf{M}}^{-1}  \left( \mathbf{A}_{1} \mathbf{Q} \mathbf{A}^{t}_{2} \right) \mathbf{H} \left( \mathbf{A}_{2} \mathbf{Q} \mathbf{A}^{t}_{1} \right) \dot{\mathbf{M}}^{-1} \mathbf{f}_{1}\\
                 = & \mathbf{f}^{t}_{2} \mathbf{H} \mathbf{f}_{2} - 2\mathbf{f}^{t}_{1} \dot{\mathbf{M}}^{-1} \bar{\mathbf{M}} \mathbf{H} \mathbf{f}_{2} + \mathbf{f}^{t}_{1} \dot{\mathbf{M}}^{-1} \bar{\mathbf{M}} \left( \mathbf{H} \bar{\mathbf{M}^{t} \dot{\mathbf{M}}^{-1}} \right) \mathbf{f}_{1}\\
        \phi_{2} = & - \mathbf{f}^{t}_{1} \dot{\mathbf{M}}^{-1} \bar{\mathbf{M}} \mathbf{G}^{t} \mathbf{f}_{1} + 2\mathbf{f}^{t}_{1} \mathbf{G} \mathbf{f}_{2} + \mathbf{f}^{t}_{2} \mathbf{H} \mathbf{f}_{2}
    \end{aligned} \label{eq:10-28}
\end{equation}
It is obvious from equations (10.25), (10.26), and (10.28) that\\
从方程\eqref{eq:10-25}、\eqref{eq:10-26}和\eqref{eq:10-28}中可以明显知道

Example 10.4. Figure 10.2 shows two adjacent plane triangles where six angles are measured.  These are uncorrelated and have the following values:\\
示例 10.4.图 10.2 显示了两个相邻平面三角形，其中测量了六个角度。 这些值不相关，具有以下值：
$$
    \begin{aligned}
        l_{1} &= 60^\circ 03'\\
        l_{2} &= 60^\circ 02'\\
        l_{3} &= 60^\circ 01'\\
        l_{4} &= 120^\circ 05'\\
        l_{5} &= 120^\circ 05'\\
        l_{6} &= 60^\circ 01'
    \end{aligned}
$$

\begin{figure}[H]
    \centering
    \includegraphics{images//10-2.png}
    \caption{ }
    \label{img:10-2}
\end{figure}
It is required to adjust the whole figure with all six observations together. \\
需要将整个图形与所有六个观测结果一起平差。

We adjust the figure in two steps: First, using angles $l_1$, $l_2$, and $l_3$, and in the second step, using the remaining data.\\
我们分两步对图形平差：第一，使用角度 $l_1$、$l_2$ 和 $l_3$，在第二步中，使用其余数据。

Then we verify the consistency of the answers from both procedures.\\
然后，我们验证两个过程的答案的一致性。

Solution\\
方案
The Direct Adjustment. \\
直接平差。

There are 2 degrees of freedom (the reader should verify this) for which we write the following two independent conditions:\\
有 2 个自由度（读者应验证这一点），为此我们编写以下两个独立的条件：
$$
    \begin{aligned}
                l_{1} + l_{2} + l_{3} - 180 &= 0\\
        l_{2} + l_{4} + l_{5} + l_{6} - 360 &= 0
    \end{aligned}
$$
Or
$$
    \left[\begin{array}{rrrrrr}
        1&1&1&0&0&0\\
        0&1&0&1&1&1
    \end{array}\right] \mathbf{v} = 
    \left[\begin{array}{l}
        180 - l_{1} - l_{2} - l_{3}\\
        360 - l_{2} - l_{4} - l_{5} - l_{6}
    \end{array}\right] = 
    \left[\begin{array}{l}
        -6"\\
        -12"
    \end{array}\right]
$$

$$
    \mathbf{k} = 
    \left[\begin{array}{rr}
        3&1\\
        1&4
    \end{array}\right]^{-1}
    \left[\begin{array}{l}
        -6\\
        -12
    \end{array}\right] = 
    \left[\begin{array}{r}
        -1.09"\\
        -2.73"
    \end{array}\right]
$$

$$
    \mathbf{v} = \mathbf{Q} \mathbf{A}^{t} \mathbf{k} = 
    \left[\begin{array}{rr}
        1&0\\
        1&1\\
        1&0\\
        0&1\\
        0&1\\
        0&1
    \end{array}\right]
    \left[\begin{array}{r}
        -1.09"\\
        -2.73
    \end{array}\right] = 
    \left[\begin{array}{r}
        -1.09"\\
        -3.82"\\
        -1.09"\\
        -2.73"\\
        -2.73"\\
        -2.73"
    \end{array}\right]
$$

$$
    \mathbf{Q}_{v v} = \frac{1}{11} 
    \left[\begin{array}{rrrrrr}
        4&3&4&-1&-1&-1\\
          5&3& 2& 2& 2\\
            4&-1&-1&-1\\
               3& 3& 3\\
                  3& 3\\
                     3
    \end{array}\right]
$$

$$
    \mathbf{Q}_{l l} = 
    \left[\begin{array}{rrrrrr}
        7&-3&-4& 1& 1& 1\\
           6&-3&-2&-2&-2\\
              7& 1& 1& 1\\
                 8&-3&-3\\
                    8&-3\\
                       8
    \end{array}\right]
$$
Adjustment in Steps\\
分步平差
First Step: Here there is only one condition,\\
第一步：这里只有一个条件，
$$
    l_{1} + l_{2} + l_{3} - 180 = 0
$$

$$
    \left[\begin{array}{rrrrrr}
        1&1&1&0&0&0
    \end{array}\right] \mathbf{v}_{1} = -6"
$$

$$
    \mathbf{k}_{1} = 2" \quad \mathbf{v}_{1} = 
    \left[\begin{array}{rrrrrr}
        -2"&-2"&-2"&0&0&0
    \end{array}\right]^{t}
$$

$$
    \mathbf{Q}_{v_{1} v_{1}} = \frac{1}{3} 
    \left[\begin{array}{rrrrrr}
        1&1&1&0&0&0\\
          1&1&0&0&0\\
            1&0&0&0\\
              0&0&0\\
                0&0\\
                  0
    \end{array}\right]
$$

$$
    \mathbf{Q} = \mathbf{Q} - \mathbf{Q}_{v_{1} v_{1}} = I - \mathbf{Q}_{v_{1} v_{1}} = \frac{1}{3} 
    \left[\begin{array}{rrrrrr}
        2&-1&-1& 0& 0& 0\\
           2&-1& 0& 0& 0\\
              2& 0& 0& 0\\
                 3& 0& 0\\
                    3& 0\\
                       3
    \end{array}\right]
$$
Second Step: The vector of observations in this step is not the one originally given at the beginning of the problem. \\
第二步：此步骤中的观测值的向量不是问题开头最初给出的向量。

It is updated by the residuals from the first step. Thus\\
它由第一步中的残差更新。因此
$$
    {l'}_{1} = 60^\circ 00'01" \quad {l'}_{2} = 60^\circ 00'00" \quad {l'}_{1} = 59^\circ 59'39"
$$

$$
    {l'}_{4} = 120^\circ 00'05" \quad {l'}_{5} = 120^\circ 00'04" \quad {l'}_{6} = 60^\circ 00'01"
$$
The one condition for this step is\\
此步骤的一个条件是
$$
    \begin{aligned}
        {l'}_{2} + {l'}_{4} + {l'}_{5} + {l'}_{6} - 360 &= 0\\
        \left[\begin{array}{rrrrrr}
            0&1&0&1&1&1
        \end{array}\right] \mathbf{v}_{2} &= 360 - {l'}_{2} - {l'}_{4} - {l'}_{5} - {l'}_{6} = -10"
    \end{aligned}
$$
Note that   is not the same as . The difference as given by equation (10.23) is    , which is correct. Thus\\
请注意，$f_{2} = -10"$与$f_{2} = -12"$不一样。方程\eqref{eq:10-23}给出的差值是$\left(-\mathbf{A}_{2} \mathbf{v}_{1} \right) = +2"$，这是正确的。因此
$$
    \begin{aligned}
        \mathbf{k}_{2} &= \left( \mathbf{A}_{2} \mathbf{Q}_{2} \mathbf{A}_{2} \right)^{-1} (-10) = -\left(\frac{30}{11}\right)"\\
        \mathbf{v}_{2} &= \mathbf{Q}_{2} \mathbf{A}_{2} \mathbf{k}_{2}\\
                       &= \left[\begin{array}{rrrrrr}
                              0.91&-1.82&0.91&-2.73&-2.73&-2.73
                          \end{array}\right]^{t} \quad \text{ ( sec of arc )}
    \end{aligned}
$$
And
$$
    \mathbf{v}_{1} + \mathbf{v}_{2} = 
    \left[\begin{array}{rrrrrr}
        -1.09&-3.82&-1.09&-2.73&-2.73&-2.73
    \end{array}\right]^{t} \quad \text{(sec of arc)}
$$

Which is identical to v computed in the direct adjustment.\\
与直接平差中计算的$\mathbf{v}$相同。
$$
    \mathbf{Q}_{v_{2} v_{2}} = \frac{1}{33} 
    \left[\begin{array}{rrrrrr}
        1&-2& 1&-3&-3&-3\\
           4&-2& 6& 6& 6\\
              1&-3&-3&-3\\
                 9& 9& 9\\
                    9& 9\\
                       9
    \end{array}\right]
$$

$$
    \bar{\mathbf{Q}}_{l l} = \mathbf{Q}_{2} - \mathbf{Q}_{v_{2} v_{2}} = \frac{1}{11} 
    \left[\begin{array}{rrrrrr}
        7& -3&-4& 1& 1& 1\\
            6&-3&-2&-2&-2\\
               7& 1& 1& 1\\
                  8&-3&-3\\
                     8&-3\\
                        8
    \end{array}\right]
$$
Which is again identical to   computed in the direct adjustment.\\
与直接平差中计算的$\mathbf{Q}_{l l}$相同。

It may be concluded that both adjustment with derived observations and adjustment in steps are feasible provided we make sure of satisfying the requirements of each concept .\\
可以得出结论，如果我们确保满足每个概念的要求，通过派生的观察值平差和分步平差都是可行的。

Most important of these are not changing the model when adjusting with derived observations and making the necessary modification of the stochastic model for both concepts of adjustment.\\
其中最重要的是，在使用派生观测值进行平差时，不会更改模型，并且对两个平差概念的随机模型进行必要的修改。

This chapter is devotes to several topics that were mentioned but not discussed in detail in the foregoing chapters. Such topics pertain mostly to practical matters of adjustment and include selection of approximations for model variables, criteria for termination of iterations for nonlinear cases, data editing for gross errors, and computational aspects of adjustment, and so on .The problem of linearization of nonlinear equations is discussed first.
